If n is an odd no. And n1 thenprove that (n, n2-1/2, n2+1/2) is a pythogoream triplet write twophythgoream triplet by substituting value of

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^{'} n^{'}

n > 1

(n, \frac{n^{2} - 1}{2}, \frac{n^{2} + 1}{2})

^{'} n^{'}

n

(n, \frac{n^{2} - 1}{2}, \frac{n^{2} + 1}{2})

n

m^{2} + {m^{'}}^{2} + 2 m m^{'} cos θ = 1

n^{2} + {n^{'}}^{2} + 2 n n^{'} cos θ = 1

m n + m^{'} n^{'} + (m n^{'} + m^{'} n) = cos θ = θ

m^{2} + n^{2} = {csc}^{2} θ

n_{1}

n_{2}

(a, b, c)

a b c = 144, (a, b, c \in N)

What is the value of n for which 8, $(n^{2} - 1)$ , 17 are Pythagorean triplet.

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